Download - C4 Chapter 1: Partial Fractions Dr J Frost ([email protected]) Last modified: 30 th August 2015
C4 Chapter 1: Partial Fractions
Dr J Frost ([email protected])www.drfrostmaths.com
Last modified: 30th August 2015
OverviewAt GCSE you learnt how to combine a sum/difference of fractions into one.We now want to learn how to do the opposite process: split a fraction into a sum of simpler ones, known as partial fractions.
1𝑥+1
+ 𝑥𝑥+2
𝑥2+2𝑥+2(𝑥+1 ) (𝑥+2 )
𝑥+1𝑥 (𝑥−1 )?
Method
Split into partial fractions.Q
METHOD 1: Substitution
If each factor of the denominator is linear, we can split like such (for constants and ):
We don’t like fractions in equations, so we could simplify this to:
METHOD 2: Equating coefficients
We can easily eliminate either or by an appropriate choice of :
If :
If :
Therefore:
Since this is an identity, the coefficients of must match, and the constant terms must match.
Solving simultaneous equations gives same solutions as before.
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Test Your UnderstandingC4 Jan 2011 Q3
Let :
Let :
Therefore Notice we can move the “–” to the front of the fraction.
Note that we don’t technically need this last line from the perspective of the mark scheme, but it’s good to just to be on the safe side
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More than two fractionsThe principle is exactly the same if we have more than two linear factors in the denominator.
Split into partial fractions.Q
When :
When :
When :
So
Bro Tip: While substitution is generally the easier method, I sometimes compare coefficients of just the term to avoid having to deal with fractions. No need to expand; we can see by observation that:
Then is easy to determine given we know and .
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Exercise 1B/1C
Express the following as partial fractions.
1
a
c
e
g
Exercise 1B
Factorise the denominator first:
Exercise 1C
1
2
a
b
c
a
c
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Repeated linear factorsSuppose we wished to express as . What’s the problem?
Because the denominators are the same, we’d get . There’s no constant values of and we can choose such that because the denominators will still be different.
Split into partial fractions.Q
11𝑥2+14 𝑥+5(𝑥+1 )2 (2𝑥+1 )
≡𝐴𝑥+1
+ 𝐵(𝑥+1 )2
+ 𝐶2𝑥+1
The problem is resolved by having the factor both squared and non-squared.
When : When :
At this point we could substitute something else (e.g. ) but it’s easier to equate terms.
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Dealing with Improper Fractions
The ‘degree’ of a polynomial is the highest power, e.g. a quadratic has degree 2.
An algebraic fraction is improper if the degree of the numerator is at least the degree of the denominator.
𝑥2−3𝑥+2
𝑥+1𝑥−1
𝑥3−𝑥2+3𝑥2−𝑥
! To split an improper fraction into partial fractions, either:1. Divide algebraically first.2. Or introduce a whole term and deal with identity immediately.
Dealing with Improper FractionsSplit into partial fractions.Q
Dividing algebraically gives:
Turn numerator back:
Let
So
Method 1: Algebraic Division Method 2: Using One Identity(method not in your textbooks but in mark schemes)
Let:
If : If : Comparing coefficients of :
Bropinion: I personally think the second method is easier. And mark schemes present it as “Method 1” – implying more standard!
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